Question: Simplify and expand the following expression: $ \dfrac{z - 6}{3z - 2}-\dfrac{3z - 9}{4z + 10} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3z - 2)(4z + 10)$ Multiply the first term by $\dfrac{4z + 10}{4z + 10}$ $ \begin{align*} \dfrac{z - 6}{3z - 2} \times \dfrac{4z + 10}{4z + 10} & = \dfrac{(z - 6)(4z + 10)}{(3z - 2)(4z + 10)} \\ & = \dfrac{4z^2 - 14z - 60}{(3z - 2)(4z + 10)}\end{align*} $ Multiply the second term by $\dfrac{3z - 2}{3z - 2}$ $ \begin{align*} \dfrac{3z - 9}{4z + 10} \times \dfrac{3z - 2}{3z - 2} & = \dfrac{(3z - 9)(3z - 2)}{(4z + 10)(3z - 2)} \\ & = \dfrac{9z^2 - 33z + 18}{(4z + 10)(3z - 2)}\end{align*} $ Now we have: $ = \dfrac{4z^2 - 14z - 60}{(3z - 2)(4z + 10)} - \dfrac{9z^2 - 33z + 18}{(4z + 10)(3z - 2)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{4z^2 - 14z - 60 - (9z^2 - 33z + 18)}{(3z - 2)(4z + 10)} $ $ = \dfrac{4z^2 - 14z - 60 - 9z^2 + 33z - 18}{(3z - 2)(4z + 10)} $ $ = \dfrac{-5z^2 + 19z - 78}{(3z - 2)(4z + 10)}$ Expand the denominator: $ = \dfrac{-5z^2 + 19z - 78}{12z^2 + 22z - 20}$